Post by thread:[EE] Killing a chip by measuring it

Quick question for you all... How real is the threat of frying a chip simply by measuring various values of its circuitry with a multimeter that's powered by a battery of a higher voltage than the chip is rated to handle? We'd built up a nice wire-wrapped circuit around an important IC and were measuring resistances and voltage differences between different points on our board, searching for potential shorts, etc., when we realized that our tiny Radio Shack multimeter was powered by a 12V battery! The chip itself operates at 3.3V and has an absolute max of 6V. Have we likely fried this thing? Seems almost Heisenberg-ian... ruining it by measuring it... Thanks for your help. Julian

Julian, Think about it. Have you measured less than 1 ohm before? 12v/1ohm = 12A !!!!!! There's likely NO WAY that 12v battery (lithium?) generated that much current, and you're circuit is just fine. The circuitry for measuring resistance is a bit more complicated than that, and it's likely the chip never saw 12v. It's VERY likely nothing's been harmed. Most of the MM have been idiot proofed and these days you can't even peg the meters. In the end, you usually don't blame a fried chip because it's almost never that (unless it got very hot, smoked, and the plastic looks melted). in other words, not cooked unless it looks toasty. Dan Crews, E.I.T. (spam_OUTcrews.danTakeThisOuTgmail.com) <><

Thanks for the response, Dan. Your e-mail confirms an internal argument of mine that I was trying to convince myself of. :) Let me be sure I understand this... Damage done to a chip is not due to the voltage applied, but the current induced, correct? If that's the case, then a chip will be rated for a maximum voltage that, given the internal impedance of that chip, a voltage beyond that rating will cause unsafe current -- IF the voltage source is able to source the expected current. In which case, a very high voltage (far beyond the rated max of the chip) won't necessarily damage the chip if the induced current isn't the "expected" result of a simple Ohm's law calculation. Right? And if this is the case, why don't the data sheets generally specify an absolute maximum current instead of a voltage? Thanks! Julian PS. Although... when I say "fried," I guess I don't necessarily mean toasted... can't the circuitry chip be damaged at a much lower current without visibly ruining the entire chip? On Sat, 12 Mar 2005 08:23:45 -0500, Dan Crews <.....crews.danKILLspam@spam@gmail.com> wrote: {Quote hidden}> Julian, > > Think about it. > > Have you measured less than 1 ohm before? > > 12v/1ohm = 12A !!!!!! > > There's likely NO WAY that 12v battery (lithium?) generated that much > current, and you're circuit is just fine. The circuitry for > measuring resistance is a bit more complicated than that, and it's > likely the chip never saw 12v. It's VERY likely nothing's been > harmed. Most of the MM have been idiot proofed and these days you > can't even peg the meters. > > In the end, you usually don't blame a fried chip because it's almost > never that (unless it got very hot, smoked, and the plastic looks > melted). in other words, not cooked unless it looks toasty. > > Dan Crews, E.I.T. > (crews.danKILLspamgmail.com) <>< > -

At 08:48 AM 3/12/2005, Senior Design wrote: >Thanks for the response, Dan. > >Your e-mail confirms an internal argument of mine that I was trying to >convince myself of. :) Let me be sure I understand this... > >Damage done to a chip is not due to the voltage applied, but the >current induced, correct? No. Mosfet gate puncture is by voltage. You leave a little crater in the gate, and then over time, conductive fingers grow through the hole in the glass due to the electrostatic attraction of the conductor atoms. You can also cook things by applying too much current.

> ... convince myself of. :) Let me be sure I understand this... > > Damage done to a chip is not due to the voltage applied, but the > current induced, correct? A little more complex than that. There are (probably) two main failure modes. 1. Power dissipated causes thermal damage 2. Voltage breaks down something which stays broken. In the latter catefory are eg FET gates where you puncture the extremely thin oxide layer. (Also Tantalum capacitors where you essentially do the same thing). As many IC's are FET based they have a very real voltage rating. You are however unlikely to get the gate voltages in the FETs to desctruction point with a multimeter as taking a pin outside the voltage range will probably cause the per-pin protection diodes to conduct and provide a *reasonably* energy resistant low impedance path. YMMV though - don't tempt fate !!! Note that protection diodes conduct at about 0.6b outside ground or Vdd rails WHEN POWER IS APPLIED but will conduct at far lower voltages when no power is applied to Vdd/ground as the voltage applied to the pin relative to ground attempts to power the IC through the protection diode. Applying voltages to pins when power is not applied can get an IC to go into funny modes because charge may go places it's not designed to go. In worst case situations, more in the good old days than now, taking a pin outside spec with power on could result int he triggering of a parasitic thyristor relative to the IC substrate and cause latchup and often enough destruction of the IC. This tends to happen far less nowadays due to suitable built in protection. SO - your chip is quite ikely OK BUT if it's not don't be totally surprised. > this is the case, why don't the data sheets generally specify an > absolute maximum current instead of a voltage? As above. Convince yourself. Get a MOSFET that you don't love too much. Find out its rated gate voltage. Calculate a resistors value that would allow energy far less than the FET should be able to stand to be dissipated at this voltage. Double the resistor value. Apply a variable voltage source gate to ground via the "safe" resistor. Exceed Vgsmax adequately and note the FET die while gate dissipation is still minimal. RM

Julian, Sorry, (egg on face) but I did oversimplify the issue quite a bit (as others have pointed out). That's the nice part of such a quality group -- they won't let the simple answers go, unless they're also dead right. Yes, voltage is a failure mode, but protection circuitry can usually keep some of it in check by trading SOME of the over-voltage for current. This shows an importance for the input's current capacity while generating that voltage. The typical active modes for a MM are VERY high impedance, which means the current capacity is very low and the voltage in the divider between input and tested resistances usually drives the tested voltage down. At that point, even with the nearly infinite input impedance to a MOS gate, the protection diodes in parallel will handle enough to keep it live. While the protection can't do much, but it saves a lot of bacon. All that being said, a MM is likely to be friendly to your circuit, but that doesn't allow for silly things like leaving MOS inputs undefined or handling them barehanded without discharging first (here in FL, we can get away with it in all but the worst part of winter. High humidity can work to your benefit). They are fragile beasts and while they can stand more than you'd imagine, they are still as breakable as glass. so in the end, you'll eventually get a feel of what will and won't damage them, and revise it as you go . . . Dan Crews, E.I.T. (.....crews.danKILLspam.....gmail.com) <><

Thanks for all of your help. I appreciate everyone's insight -- I guess we'll just wait until our design is done at this point and see how things turned out with the IC. In the mean time, I'm laying off the 12V multimeter... Thanks again. Julian On Mon, 14 Mar 2005 08:15:38 -0500, Dan Crews <EraseMEcrews.danspam_OUTTakeThisOuTgmail.com> wrote: {Quote hidden}> Julian, > > Sorry, (egg on face) but I did oversimplify the issue quite a bit (as > others have pointed out). That's the nice part of such a quality > group -- they won't let the simple answers go, unless they're also > dead right. > > Yes, voltage is a failure mode, but protection circuitry can usually > keep some of it in check by trading SOME of the over-voltage for > current. This shows an importance for the input's current capacity > while generating that voltage. The typical active modes for a MM are > VERY high impedance, which means the current capacity is very low and > the voltage in the divider between input and tested resistances > usually drives the tested voltage down. At that point, even with the > nearly infinite input impedance to a MOS gate, the protection diodes > in parallel will handle enough to keep it live. While the protection > can't do much, but it saves a lot of bacon. > > All that being said, a MM is likely to be friendly to your circuit, > but that doesn't allow for silly things like leaving MOS inputs > undefined or handling them barehanded without discharging first (here > in FL, we can get away with it in all but the worst part of winter. > High humidity can work to your benefit). They are fragile beasts and > while they can stand more than you'd imagine, they are still as > breakable as glass. > > so in the end, you'll eventually get a feel of what will and won't > damage them, and revise it as you go . . . > > > Dan Crews, E.I.T. > (crews.danspam_OUTgmail.com) <>< > -